Question

A hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3x^2$ $+4y^2=12,$ then this hyperbola does not pass through which of the following points?

• A. $\left(1,-\frac{1}{\sqrt{2}}\right)$
• B. $\left(\sqrt{\frac{3}{2}},\frac{1}{\sqrt{2}}\right)$
• C. $\left(\frac{1}{\sqrt{2}},0\right)$
• D. $\left(-\sqrt{\frac{3}{2}},1\right)$

Answer 1

Answer: (B)

Ellipse $:\frac{x^2}{4}+\frac{y^2}{3}=1$
eccentricity $=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
$\therefore$ foci $=(\pm 1,0)$
for hyperbola, given $2a=\sqrt{2}$
$\Rightarrow a=\frac{1}{\sqrt{2}}$
$\therefore$ hyperbola will be
$\frac{x^2}{1/2}-\frac{y^2}{b^2}=1$
eccentricity $=\sqrt{1+2b^2}$
$\therefore$ foci $=\left(\pm \sqrt{\frac{1+2b^2}{2}},0\right)$
$\because$ Ellipse and hyperbola have same foci
$\Rightarrow \sqrt{\frac{1+2b^2}{2}}=1$
$\Rightarrow b^2=\frac{1}{2}$
$\therefore$ Equation of hyperbola :
$\frac{x^2}{1/2}-\frac{y^2}{1/2}=1$
$\Rightarrow x^2-y^2=\frac{1}{2}$
Clearly $\left(\sqrt{\frac{3}{2}},\frac{1}{\sqrt{2}}\right)$ does not lie on it.